Question: How many of the first $1000$ positive integers can be expressed in the form
\[\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor\]where $x$ is a real number, and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$?
Explanation: Let $f(x)$ be the given expression. We first examine the possible values of $f(x)$ for $x$ in the interval $(0, 1].$ Note that $f(0) = 0,$ while $f(1) = 2 + 4 + 6 + 8 = 20.$

As we increase $x$ from $0$ to $1,$ each of the four floor functions "jumps up" by $1$ at certain points. Furthermore, if multiple floor functions "jump up" at the same value of $x,$ then some integers will be skipped.

For each $k,$ the function $\lfloor kx \rfloor$ "jumps up" at $x = \tfrac{1}{k}, \tfrac{2}{k}, \ldots, \tfrac{k-1}{k}, \tfrac{k}{k}.$ Therefore, we see that at $x = \tfrac{1}{2}$ and $x = 1,$ all four of the given functions "jump up," so that three integers are skipped. Also, for $x = \tfrac{1}{4}$ and $x =\tfrac{3}{4},$ the functions $\lfloor 4x \rfloor$ and $\lfloor 8x \rfloor$ both "jump up," skipping one integer.

Thus, for $0 < x \le 1,$ $f(x)$ takes $20 - 3 - 3 - 1 - 1 = 12$ positive integer values. Notice that \[\begin{aligned} f(x+1) &= \lfloor 2(x+1) \rfloor + \lfloor 4(x+1) \rfloor + \lfloor 6(x+1) \rfloor + \lfloor 8(x+1) \rfloor \\ &= \left(\lfloor 2x \rfloor+2\right) + \left(\lfloor 4x \rfloor +4\right)+ \left(\lfloor 6x\rfloor+6 \right)+ \left(\lfloor 8x \rfloor +8\right) \\ &= f(x) + 20. \end{aligned}\]Therefore, in the interval $1 < x \le 2,$ $f(x)$ takes $12$ more integer values between $21$ and $40,$ respectively. In general, $f(x)$ takes $12$ out of every $20$ positive integer values from the list $20a, 20a+1, \ldots, 2a+19.$

Since $20$ is a divisor of $1000,$ exactly $\tfrac{12}{20} = \tfrac{3}{5}$ of the first $1000$ positive integers are possible values for $f(x).$ Thus the answer is $1000 \cdot \tfrac{3}{5} = \boxed{600}.$